Question

The physical properties of the relaxed eye are: distance from front surface of cornea to front surface of the lens: 3.6 mm, thickness of lens: 3.6 mm, cornea (r1): 8 mm, anterior surface of lens (r2):10.0 mm, posterior surface of lens (r3): 6.0 mm. The refractive index of aqueous and vitreous humor and lens are 1.34 and 1.41, respectively. With these properties, the distance from the cornea to the retina is 25 mm and an object infinitely far away is in focus on the retina. An object is placed a distance of 1.0 m away from the cornea.

The eye accommodates by changing the radius of curvature of the posterior surface of the lens. Find the new radius of curvature. Assume that the changes are small enough that the distances between the surfaces remain unchanged.

The solution for the curvature is -5.4 mm. I am unsure how the value is obtained. Thanks!

Answer 1

I did not use the given distance between cornea and retina(yields slightly incorrect value); rather, I calculated the position where the image forms when the object is at infinity and then wrote equations where the final position would remain the same.